Theorem 32.2.2.label Let $G$ be a locally compact group, $(H, \pi_{0})$ be a unitary representation of $G$, and $\pi: L^{1}(G; \complex) \to B(H)$ be the *-representation determined by $\pi_{0}$, then the von Neumann algebras generated by $\pi_{0}(G)$ and $\pi(L^{1}(G; \complex))$ coincide.
Proof. By (3) of Definition 32.2.1, $\pi_{0}(G)$ lies in the strong-operator closure of $\pi(L^{1}(G; \complex))$.
On the other hand, let $K \subset G$ be compact, $\eps > 0$, and $\seqf{(\xi_j, \eta_j)}\subset H^{2}$, then there exist non-empty sets $\bracsn{V_k}_{1}^{m} \subset 2^{G}$ such that $K = \bigsqcup_{k = 1}^{m} V_{k}$ and $|\dpn{[\pi_0(x) - \pi_0(y)]\xi_j, \eta_j}{H}| \le \eps$ for all $x, y \in V_{k}$, $1 \le j \le n$, and $1 \le k \le m$.
Let $\seqf{x_k}\subset G$ with $x_{k} \in V_{k}$ for all $1 \le k \le m$, then
so $\pi(\one_{K})$ is in the von Neumann algebra generated by $\pi_{0}(G)$. By inner regularity of the Haar measure, $\pi(\one_{A})$ is in the von Neumann algebra generated by $\pi_{0}(G)$ for all $A \in \cb_{G}$ with $\mu(A) < \infty$. By density of simple functions, $\pi(L^{1}(G; \complex))$ is in the von Neumann algebra generated by $\pi_{0}(G)$.$\square$
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